Question

given the table below write the equation of the quadratic in vertex form then convert it to standard form.

x | -5 | -4 | -3 | -2 | -1
f(x) | -1 | -4 | -5 | -4 | -1

vertex form:
*write the equation in the form y=a(x-h)^2+k without spaces.

standard form:
*write the equation in the form y=ax^2+bx+c without spaces.

Explanation:

Step1: Find the vertex

The vertex of a parabola is the midpoint between symmetric points. Looking at the table, when \( x = -5 \) and \( x = -1 \), \( f(x) = -1 \); when \( x = -4 \) and \( x = -2 \), \( f(x) = -4 \). The axis of symmetry (x - coordinate of vertex) is the average of \( -5 \) and \( -1 \) (or \( -4 \) and \( -2 \)): \( h=\frac{-5 + (-1)}{2}=\frac{-6}{2}=-3 \). The y - coordinate of the vertex \( k=-5 \) (from \( x = -3 \), \( f(x)=-5 \)). So the vertex form is \( y = a(x + 3)^{2}-5 \).

Step2: Find the value of a

We can use a point from the table, for example, when \( x=-5 \), \( f(x)=-1 \). Substitute into the vertex form:
\( -1=a(-5 + 3)^{2}-5 \)
\( -1=a(-2)^{2}-5 \)
\( -1 = 4a-5 \)
Add 5 to both sides: \( 4=4a \)
Divide both sides by 4: \( a = 1 \)

Step3: Write the vertex form

Substitute \( a = 1 \), \( h=-3 \), \( k = -5 \) into \( y=a(x - h)^{2}+k \):
\( y=(x + 3)^{2}-5 \)

Step4: Convert to standard form

Expand \( (x + 3)^{2}-5 \):
First, \( (x + 3)^{2}=x^{2}+6x + 9 \)
Then, \( x^{2}+6x + 9-5=x^{2}+6x + 4 \)

Answer:

Vertex Form: \( y=(x + 3)^{2}-5 \)
Standard Form: \( y=x^{2}+6x + 4 \)